# integration part 2

• Feb 25th 2008, 01:45 PM
jliddell
integration part 2
$\displaystyle \int (xsinx^2cosx^2),dx$

all i need are steps leading to the answer
• Feb 25th 2008, 01:46 PM
Krizalid
Start by makin' $\displaystyle u=x^2.$
• Feb 25th 2008, 01:58 PM
jliddell
integration part 2
ok not quite getting it. can you help me out a little more please.
• Feb 25th 2008, 02:00 PM
Krizalid
$\displaystyle u=x^2\implies du=2x\,dx,$

$\displaystyle \int {x\sin x^2 \cos x^2 \,dx} = \frac{1} {2}\int {\sin u\cos u\,du} = \frac{1} {4}\int {\sin 2u\,du} .$
• Feb 26th 2008, 10:57 AM
simsima_1
needing a little more help w/ this....
ok so the answer is

$\displaystyle (x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C$

and i'm not sure how i would get that using

$\displaystyle {1/4} \int sin2u$
• Feb 26th 2008, 11:20 AM
Jhevon
Quote:

Originally Posted by simsima_1
ok so the answer is

$\displaystyle (x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C$

and i'm not sure how i would get that using

$\displaystyle {1/4} \int sin2u$

there are often many different forms for the answer to an integral. following Krizalid's suggestion is good enough.